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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    Mean value theorem states that for continuous function f on [a, b] and differentiable on (a, b), there is at least one point c \in (a, b) such that f'(c) = \frac{f(b) - f(a)}{b-a}.

    Consider the function f(x) = x^3.
    f'(x) = 0 at x = 0. However, there is no interval [a , b] that satisfies mean value theorem (since f(b) > f(a) for all b > a).

    Why does the Mean value theorem fail here?
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  2. #2
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    Quote Originally Posted by Rozaline View Post
    Mean value theorem states that for continuous function f on [a, b] and differentiable on (a, b), there is at least one point c \in (a, b) such that f'(c) = \frac{f(b) - f(a)}{b-a}.

    Consider the function f(x) = x^3.
    f'(x) = 0 at x = 0. However, there is no interval [a , b] that satisfies mean value theorem (since f(b) > f(a) for all b > a).

    Why does the Mean value theorem fail here?

    It doesn't: the logical direction is ==> , not <== : for any a < b in the interval we have [f(b) - f(a)]/[b-a] = f'(c), for some c in (a,b), but this doesn't meant that ANY value f'(c) of the derivative has to be obtained this way.

    Tonio
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